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Lorentz transformations: how can I derive the time equation?

  1. Oct 8, 2012 #1
    Starting with:

    x'=[itex]\gamma[/itex](x-vt) & x=[itex]\gamma[/itex](x'+vt')

    I know that I can derive t'=[itex]\gamma[/itex](t-vx/c^2)... however I can't seem to make it fall out mathematically. The suggested method is to cancel x'. Can anyone help me out on the steps?

    Much appreciated! :)
     
  2. jcsd
  3. Oct 8, 2012 #2

    jtbell

    User Avatar

    Staff: Mentor

    Show us how far you've gotten, and maybe someone will give you a nudge from there.
     
  4. Oct 8, 2012 #3


    Here is a derivation of time dilation based on the Pythagorean Theorem. It begins with the proposition that all observers measure the same value for the speed of light, regardless of their relative speeds. For this to be true, one picture of such a universe logically includes four dimensions; different observers moving at different constant relativistic speeds relative to each other would be associated with 4-dimensional worldlines slanted with respect to an arbitrarily selected rest frame. An observer's worldline is colinear with his X4 axis along the 4th dimension ("time axis"). Further, the X1 axis is slanted as well, such that the 45-degree photon world line always bisects the angle between X1 and X4 (this assures the same measurement of "c" in all inertial frames)--that's just the way the universe is constructed in four dimensions. In the sketch below, the X1 axes represent the cross-section of the 4-dimensional universe (X2 and X3 are suppressed for ease of viewing) experienced by an observer at some instant of time (some point along the X4 dimension). This sets up the geometry in which you apply the Pythagorean Theorem. In this sketch a blue guy and red guy move in opposite directions at the same relativistic speed relative to the black rest frame. We regard any observer to be moving along his own X4 axis at the speed of light: X4 = ct, or t = X4/c
    Four_dimensional_Space.jpg [/QUOTE]
     
    Last edited: Oct 8, 2012
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